Question

Liquid hexane
(CH,(CH), CH) will react with gaseous oxygen (0) to produce gaseous carbon dioxide (CO2) and gaseous water (1,0). Suppose 1.72 g
of hexane is mixed with 8.0 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the
correct number of significant digits.

Answer 1

Answer: The mass of $H_2O$ produced is 2.52 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

• For hexane:

Given mass of hexane = 1.72 g

Molar mass of hexane = 86.18 g/mol

Putting values in equation 1, we get:

$\text{Moles of hexane}=\frac{1.72g}{86.18g/mol}=0.020mol$

• For oxygen gas:

Given mass of oxygen gas = 8.0 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{8.0g}{32g/mol}=0.25mol$

The chemical equation for the combustion of hexane follows:

$2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O$

By stoichiometry of the reaction:

If 2 moles of hexane reacts with 19 moles of oxygen gas

So, 0.020 moles of hexane will react with = $\frac{19}{2}\times 0.020=0.19mol$ of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hexane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of hexane produces 14 moles of $H_2O$

So, 0.020 moles of hexane will produce = $\frac{14}{2}\times 0.020=0.14mol$ of $H_2O$

We know, molar mass of $H_2O$ = 18 g/mol

Putting values in above equation, we get:

$\text{Mass of }H_2O=(0.14mol\times 18g/mol)=2.52g$

Hence, the mass of $H_2O$ produced is 2.52 g